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With the nullish coalescing operator (??), one is tempted to use it for specifying defaults for non-existent object keys, as suggested in an answer here, e.g.:

let a = foo.bar ?? 'default';

However, it's alternatives in this case (mainly in and hasOwnProperty) have their caveats and risks. Calling <key> in <object> matches keys in an object's prototype chain, as discussed here. Calling <object>.hasOwnProperty(<key>) won't work, for example, if the object's hasOwnProperty is shadowed, as discussed here. Surely, ?? has its own gotchas to watch out for.

What are the pitfalls with ?? compared to the other methods of checking for an existence of a key in any object?

In other words, are the risks of doing the following:

let a = foo.bar ?? 'default';

Closer to this

let a = 'bar' in foo ? foo.bar : 'default';

Or this

let a = foo.hasOwnProperty('bar') ? foo.bar : 'default';

Or this

let a = Object.prototype.hasOwnProperty.call(foo, 'bar') ? foo.bar : 'default';

Or are they different from either?

vasilyrud
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    `??` is not purposed to check the existence, it's purposed to check a value. – Teemu Jan 14 '22 at 13:33
  • If `foo` is undefined then `a = foo.bar ?? 'default'` will throw an error – evolutionxbox Jan 14 '22 at 13:34
  • If you're unsure of `hasOwnProperty`, you can do `Object.getOwnPropertyDescriptor(obj, 'prop');`, that method is less likely overridden. That provides `obj` to exist, though. – Teemu Jan 14 '22 at 13:44

1 Answers1

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The risks you are talking about are irrelevant when you know what kind of object to expect, in your particular case one with a bar property. There is no Object.prototype.bar.

The "risks" of using ?? after a property access are exactly the same as the risks of doing the property access. Your statement let a = foo.bar ?? 'default'; is equivalent to

let a = foo.bar == null ? 'default' : foo.bar;

or (without accessing the property twice)

let a = foo.bar;
if (a == null) a = 'default';
Bergi
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